Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
//package  com.company;
import java.io.*;
import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner s=new Scanner(System.in);
        int n=s.nextInt();
        int m=s.nextInt();
        int[][] a=new int[n][m];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                a[i][j]=s.nextInt();
            }
        }StringBuilder sb=new StringBuilder();int flag=0;int[] c=new int[m];int min=Integer.MAX_VALUE;int l=0;
        if(n<m){for(int i=0;i<n;i++){min=Integer.MAX_VALUE;
            for(int j=0;j<m;j++){
                min=Math.min(a[i][j]-c[j],min);

            }
            for(int j=1;j<=min;j++){l++;
                sb.append("row "+(i+1)+"\n");}
            for(int j=0;j<m;j++){
                if(a[i][j]-c[j]>min&&i!=0){
                   // System.out.println(i+"  "+j+" "+c[j]);
                    System.out.println(-1);flag=1;break;
                }                else if(a[i][j]-c[j]>min&&i==0){
                    c[j]=a[i][j]-min;
                   for(int k=1;k<=c[j];k++){l++;
                    sb.append("col "+(j+1)+"\n");}


                }

            }


            if(flag==1) break;   }}
        else{ c=new int[n];
            for(int i=0;i<m;i++){min=Integer.MAX_VALUE;
                for(int j=0;j<n;j++){
                    min=Math.min(a[j][i]-c[j],min);
                    }
                for(int j=1;j<=min;j++){l++;
                    sb.append("col "+(i+1)+"\n");}
                for(int j=0;j<n;j++){
                    if(a[j][i]-c[j]>min&&i!=0){
                        // System.out.println(i+"  "+j+" "+c[j]);
                        System.out.println(-1);flag=1;break;
                    }                else if(a[j][i]-c[j]>min&&i==0){
                        c[j]=a[j][i]-min;
                        for(int k=1;k<=c[j];k++){l++;
                            sb.append("row "+(j+1)+"\n");}


                    }

                }


                if(flag==1) break;}
        }


        if(flag==0){
            System.out.println(l);
            System.out.println(sb.toString());
        }

    }
}